Lenore Flo oxImate Metnods PMPE and PVD provide alternative formulation for probus in structured mechanics. However Hey dont te us how to obtain Hre soltion, just sowe conditions tort the soltion Must saTis kem过pprt (usu elly Inea stions the unknown sowton fields are replaced wi ith e lines conbination of fonctions of sowed functioned dependence. The wn &nouns of te sAten are the promoters appearing in the linear cowloination ot tuctas The resulting lineor combination wlth parameters determina frow the soltion of the algebrac systen is an approxiwotan to the exad sdotiovn. Snce the exaa sowtiondsrnot in eneral De repressed simple functions an error is introduced Rayeich-Rit+t Mehod Introduce a lex Combination of functions of given functiond formas He epproxionte soltion of our ed火() 阅)(分)=建)我=4
We have effectively replaced our "infinite dimneusiond broblem of deterlnine ai()with the problem determining the 3x n coefficints] or paremeters 收 owers this end we introduce out epproxietcon (Xa)in the definition of the poten til energ t an eortc boay「8e 个((am((2)=m(c k x his resols in an epression whose only unknows are ne Coefficients ck. These, are obtained by invokin Hne PMpE A necessary condition for is is the ALL the derivatives of the appan. this furnishes BxnI to its pareMeters should vani eQustions: kuz earths: e valder he potent id enery寸w含正Wse erdf u:ds 从2
=Acsk4E:N-套5-(:u以一 2 w动ta,41 ≥51-4(3+)=4厂肉+奇 ()-(A-(c国④+刻 (出-:出A And this gets reslly messy when exp anded. the torit thing to note ts thef because of Hwe Quadratic dependence of the interne energy on Eii, the epproxiwete strah enery wiw here terms tht ehe surdreftic in the anknown s. The_poential f the exter ne forces only has linear depadhe on Ss When le dordives 4s ore coot ed we obtain, a war aco lineer systen on he c knowns indri This corresponds i the Qener d 3D poolewe We wil L ws pocedure to specht problems (beens for 过+Fes甲pn
E a:解5be nifon distr bored load是 L工 T()=(E()+上8址()dk 个(④)2A(边个k+(9)c业 8T=o当T(cA)O I ck_A2EI C下8x+ +(q()20动x ace E工 dxal c dx 2 dx K x R xA tix vedor Dx1 vedor
⑤ 2d坠: L (L-X) ore th sahs entia Bo 。(-。最女一k一 kxk4L-(k+y ￠=最(小)Lx2(2 ck(x业=K数 d 8Q9(句=Q2( uniform loe =2)=x(-)=×2(-x) 1。-249-2L-6x K. EI(d dx,[EI( 2dx- 4EIL K E E工
K2=E(4)+E工22=2Er2 Ka2=Ed=((4=4E工 =×91 R2=2 12 2E1[42,)=2 2L42C CC 0(x)==9X1+8C 94E工 leT
